Question

)* (Vmax). 4. When [S] = KM, Vo=( A) 0.5 B) KM C) 0.75 D) kcat E) [S] 5. The overall transformation shown in the following reaction: E s. Es p + E For the reaction, the steady state assumption A) ES breakdown occurs at the same rate as ES formation B) [P]>>[E] C) implies that ki-k-1 D) implies that k., and k2 are such that the [ES] = k1[ES] E) [S] = [P]

Answer 1

NELL Mechanism: S + E THESE SE K2 PTE The enzyme combines with substrate to form a substrate enzyme complex. which may be either deactivated on can be transformed into product. In this case the nate of deactivation of the substrate enzyme complex into substrate and enzyme much much less than the formation of pono duet. ie, ką [SE] » Ky [SE] A K2 »k, ie, the complex is a ranthoff complex and we can apply steady state approximation for SE dise) = k, ts] [E] - Ky [SE] -k, [SE] =0 It is clean that, [E] = [E]. - [ES] whene, [9 = conc. of the free enzyme [E]. = conc. A the enzyme mitially [BE ] = conc. of the enzyme which is complete

Then, md ki[s] {[!). - [ES]}-K4 [SE] - K, ISUJ = 0 > Kits) [1]. – Sky [s] + K-1 +Kz} [st] = > {k, [s]+k++ kz | (st) = Ky Es] [E]. SE) = K [52[1]. K1+k₂ tk, ts] SE] = UTE. K-1 tkz + [s] ki [SE] = [[E). km + [5) where km = -K the as = Michaeles constant. 20 - 2013 ki [56] kz [s][!). KMT [s] RSDr. KM + [5] -me = kx + 3 tae-T when the conc. of substrat is very low ic, ts] « km KM+ TS] 2 km Then, ne = hankem 1s] [t]. - Kelf ts] This is effective 1st onder von w.nit substrate as [Elo is constant

case-2 when the cone. of substrate is very huge. ie, [5 km Then, [s]+kme [s] ne = K2 (5) TE). = K2 LET = constant. remax = K₂ [E lo this is effective teno rondens sxn wanit substrate conc This is the maximum velouty of the non ton a given conc. of enzyme. max 2e7 remax → [S] when [5] = km ze = Yz K2 [E]. = % remax

Amwer: 4) When [3] = km. Vo = 2 Vmax [vo =0.5&max option ④ 5 At steady state, dtes] _ at to is rate of formation of Es = hate of breakdown of Es. - option ®